arc-shaped light streak in sky representing methods of integration

Calculus

Methods Of Integration: Techniques & Examples

08.18.2022 • 8 min read

Rachel McLean

Subject Matter Expert

In this article, we review the definition of the integral. We discuss the most important integration rules to know and apply these rules to four primary methods of integration. Using examples, we practice integrating these four techniques.

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In This Article

  1. What Is Integration in Calculus?

  2. 9 Essential Integration Rules

  3. 4 Types of Integration Methods

  4. More Calculus Practice

What Is Integration in Calculus?

Differentiation finds the instantaneous rate of change of a function. Integration finds the reverse of a derivative, called the antiderivative. Integration can be conceptualized as the reverse process of differentiation.

Together, differentiation and integration constitute the fundamental operations of calculus. Their relationship is explained by the fundamental theorem of calculus.

There are two types of integrals: indefinite integrals and definite integrals. Indefinite integrals find antiderivative functions. Definite integrals calculate areas under the curve on a specific interval.

The notation for an indefinite integral is given below:

f(x)dx=F(x)+C\int f(x)dx = F(x) + C

An indefinite integral finds the antiderivatives of f(x)f(x), which are usually denoted by F(x)F(x). Taking the derivative of F(x)F(x) outputs back f(x)f(x), which explains why FF is called an antiderivative function. For this reason, integration is also sometimes referred to as anti-differentiation:

Iff(x)dx=F(x)+C, then F(x)=f(x).\text{If} \int f(x)\,dx = F(x) + C, \text{ then } F’(x) = f(x).

For example,

Graph showing integration, differentiation, and relationship guaranteed by the First Fundamental Theorem of Calculus

The symbol \int is called the integral sign, and f(x)f(x) is called the integrand. The capital letter CC is called the constant of integration. CC is used to represent any constant value.

Remember that taking the derivative of the antiderivative of f(x)f(x) gives back f(x)f(x) itself. No matter what constant value CC holds, the derivative of CC will always be zero. So, when we take the derivative of F(x)+CF(x) + C, we always get back f(x)f(x) alone. Thus CC can take any value, and for this reason, the notation F(x)+CF(x) + C represents an infinitely large family of all antiderivatives of the function ff.

The first fundamental theorem of calculus guarantees the relationship described earlier. If ff is a continuous function on an interval containing aa, then we can define FF by:

F(x)=axf(t)dtF(x) = \int_a^x f(t)\,dt

Then F(x)=f(x)F’(x) = f(x) and so F(x)F(x) is an antiderivative of f(x)f(x):

ddxaxf(t)dt=F(x)=f(x)\frac{d}{dx} \int_a^x f(t)\,dt = F'(x) = f(x)

Now, let’s review the notation for a definite integral:

abf(x)dx=A\int_{a}^{b} f(x)\,dx = A
Graph showing Second Fundamental Theorem of Calculus

AA represents the area under the curve on [a,b][a, b]. To approximate this area, we could create nn rectangles of equal width that approximate the curve, and then take the sum of their areas. This approximation is called a Riemann sum. However, this sum either overestimates or underestimates the area under the curve.

By taking the limit of the Riemann sum as the number of subdivisions nn approaches infinity, we obtain the precise area under the curve. This defines a definite integral.

The second fundamental theorem of calculus clarifies the relationship between the integral and the antiderivative function. This theorem tells us that we can find the definite integral of a function on [a,b][a, b] by taking the difference between the indefinite integral of the function evaluated at aa and the indefinite integral of the function evaluated at bb.

abf(x)dx=F(x)ab=F(b)F(a)\int_{a}^{b} f(x)\,dx = F(x)\Big|_a^b = F(b) - F(a)

For a good overview of this theorem, Dr. Hannah Fry shares what this theorem actually means, how it’s calculated, and what it can further allow us to do:

9 Essential Integration Rules

Now that we’ve defined integrals, it’s important to know how to evaluate them. These integration formulas are used in every method of integration, so it’s a good idea to memorize them.

1. Sum Rule

[f(x)+g(x)]dx=f(x)dx+g(x)dx\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx

2. Difference Rule

[f(x)g(x)]dx=f(x)dxg(x)dx\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx

3. Constant Multiplier Rule

kf(x)dx=kf(x)dx\int kf(x)\,dx = k\int f(x)\,dx for some constant kk

4. Power Rule

xndx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C for some real number nn

5. Constant Rule

adx=ax+C\int a\,dx = ax + C for some constant aa

6. Reciprocal Rules

  • 1xdx=x1dx=lnx+C\int \frac{1}{x}\,dx = \int x^{-1}\,dx = \ln{|x|} + C

  • 1ax+bdx=1aln(ax+b)+C\int\frac{1}{ax+b}\,dx = \frac{1}{a} \ln{(ax+b)} + C

Dr. Tim Chartier covers core differentiation rules and examples of how to use them:

7. Exponential Functions & Logarithmic Functions Rules

  • exdx=ex+C\int e^x\,dx = e^x + C

  • axdx=axln(a)+C\int a^x\,dx = \frac{a^x}{\ln{(a)}} + C, for any positive real number aa

  • ln(x)dx=xln(x)x+C\int \ln{(x)}\,dx = x\ln{(x)}-x + C

8. Trigonometric Functions & Inverse Trigonometric Functions Rules

For these rules, assume that xx is in radians.

  • sinxdx=cosx+C\int \sin{x}\,dx = -\cos{x} + C

  • cosxdx=sinx+C\int \cos{x}\,dx = \sin{x} + C

  • sec2xdx=tanx+C\int \sec ^2 x\,dx = \tan{x} + C

  • csc2xdx=cotx+C\int \csc ^2 x\,dx = -\cot{x} + C

  • secxtanxdx=secx+C\int \sec{x}\tan{x}\,dx = \sec{x} + C

  • cscxcotxdx=cscx+C\int \csc{x}\cot{x}\,dx = -\csc{x} + C

  • dx1x2=sin1x+C\int \frac{dx}{\sqrt{1-x^2}}=\sin ^{-1}x + C

  • dx1x2=cos1x+C\int \frac{-dx}{\sqrt{1-x^2}}=\cos ^{-1}x + C

  • dx1+x2=tan1x+C\int \frac{dx}{1+x^2}=\tan ^{-1}x + C

  • sin(ax)dx=cos(ax)a+C\int \sin{(ax)}\,dx = \frac{-\cos{(ax)}}{a}+C for some real number aa

  • cos(ax)dx=sin(ax)a+C\int \cos{(ax)}\,dx = \frac{\sin{(ax)}}{a}+C for some real number aa

9. Absolute Value Rule

xdx=xx2+C\int |x|\,dx = \frac{x |x|}{2} + C

4 Types of Integration Methods

We’ll discuss four essential techniques of integration:

  1. Integration by u-substitution

  2. Integration by parts

  3. Integration by partial fractions

  4. Integration using trigonometric identities

Integration by U-Substitution

U-substitution is used to integrate composite functions. U-Substitution reverses the chain rule. For this reason, u-substitution is also called the reverse chain rule. To use u-substitution, we rewrite our integral in terms of uu and dudu:

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)\,dx = \int f(u)\,du

We can use the four steps below for integration by substitution:

  1. Pick one part of the integrand to represent uu. Usually, uu is equal to g(x)g(x), the inner function of the composite function.

  2. Differentiate uu to find dudu. If necessary, rearrange the problem algebraically so that dudu perfectly matches what’s left inside the integral.

  3. Substitute uu and dudu into the integrand, and integrate. If taking a definite integral, redefine your limits of integration in terms of uu.

  4. If taking an indefinite integral, substitute the original values back into the resulting function and add the constant of integration to your final answer. If taking a definite integral, simply evaluate using the new limits in terms of uu.

For example, let’s evaluate [ln(x)]2xdx\int \frac{[\ln{(x)}]^2}{x}dx. Let u=ln(x)u = \ln{(x)}, so that du=1xdxdu = \frac{1}{x}dx. Substituting uu and dudu into the integrand, we get:

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)dx = \int f(u)du
[ln(x)]2xdx=u2du\int \frac{[\ln{(x)}]^2}{x}dx = \int u^2 du
=u33= \frac{u^3}{3}
=[ln(x)]33+C= \frac{[\ln{(x)}]^3}{3} + C

Integration by Parts

Integration by parts is used to take the integral of a product of functions. Integration by parts uses the formula below, which is derived directly from the product rule for derivatives:

udv=uvvdu\int udv = uv - \int vdu

We can use the four steps below to integrate by parts:

  1. Choose uu and dvdv to separate the given function into a product of functions. Generally, uu is the term that’s easiest to differentiate, and dvdv is the term that’s easiest to integrate.

  2. Differentiate uu to find dudu, and integrate dvdv to find vv.

  3. Plug uu, vv, and dudu into the integration by parts formula.

  4. Solve and simplify where needed.

For example, let’s evaluate xsin(x)dx\int x\sin(x)dx. Let u=xu=x and dv=sin(x)dxdv=\sin(x)dx. Differentiating uu, we find that du=1dxdu = 1dx. Integrating dvdv, we find that v=cos(x)v = - \cos(x). Plugging u=xu=x, v=cos(x)v = - \cos(x), du=1dxdu = 1dx, and dv=sin(x)dxdv=\sin(x)dx into the integration by parts formula, we get:

udv=uvvdu\int udv = uv - \int vdu
xsin(x)dx=xcos(x)cos(x)dx\int x\sin(x)dx = -x \cos(x) - \int - \cos(x)dx
xsin(x)dx=xcos(x)+sin(x)+C\int x\sin(x)dx = -x \cos(x) + \sin(x) + C

The acronym LIATE can help you decide which term to designate as uu. LIATE stands for:

  • Logarithmic functions

  • Inverse trigonometric functions

  • Algebraic functions

  • Trigonometric functions

  • Exponential functions

Function types that sit the highest on this list should be prioritized as uu.

Integration by Partial Fractions

Integration by partial fractions is used to integrate rational functions. A rational function is an algebraic fraction where both the numerator and denominator are polynomials.

We can use the nine steps below to integrate by partial fractions:

  1. Factor the denominator of the function.

  2. Decompose the function into a sum of its parts by assigning an unknown variable to each term of the denominator.

  3. Combine all terms into one by finding the common denominator, making sure you multiply each numerator appropriately.

  4. Multiply out the numerator.

  5. Set up an equation that equates the xx terms of the original function’s numerator with the xx terms in your new equation’s numerator.

  6. Set up a second equation that equates the constant terms of the original function’s numerator with the constant terms in your new numerator.

  7. Solve for the unknown variables.

  8. Plug your solved variables into Step 2.

  9. Integrate using the reciprocal rules.

For example, let’s evaluate x+4x2+x12dx\int \frac{x+4}{x^2+x-12}dx. First, we’ll factor the denominator. Then, we can proceed with steps 2 through 4.

x+4x2+x12dx=x+4(x+4)(x3)dx\int \frac{x+4}{x^2+x-12}\,dx = \int \frac{x+4}{(x+4)(x-3)}\,dx

Now,

x+4(x+4)(x3)=Ax+4+Bx3\frac{x+4}{(x+4)(x-3)} = \frac{A}{x+4} + \frac{B}{x-3}
=A(x3)+B(x+4)(x+4)(x3)=\frac{A(x-3) + B(x+4)}{(x+4)(x-3)}
=Ax3A+Bx+4B(x+4)(x3)=\frac{Ax-3A+Bx+4B}{(x+4)(x-3)}

Our next step is to solve for AA and BB using steps 5 through 7. Our first equation is Ax+Bx=xAx + Bx = x. We can simplify this equation to A+B=1A+B=1. Our second equation is 3A+4B=4-3A+4B=4. This gives us a system of equations, which outputs that A=0A = 0 and B=1B = 1.

Using steps 8 through 9, we have:

x+4(x+4)(x3)dx=0x+4dx+1x3dx\int \frac{x+4}{(x+4)(x-3)}\,dx = \int \frac{0}{x+4}dx + \frac{1}{x-3}\,dx
=ln(x3)= \ln{(x-3)}

This technique is also called partial fraction decomposition.

Integration Using Trigonometric Identities

Here are some of the most important trigonometric identities to know:

Pythagorean Identities

  • sin2x+cos2x=1\sin^2x + \cos^2x = 1

  • sec2xtan2x=1\sec^2 x - \tan^2 x = 1

  • csc2xcot2x=1\csc^2 x - \cot^2 x = 1

Quotient and Reciprocal Identities

  • tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}

  • cotx=cosxsinx\cot x = \frac{\cos x}{\sin x}

  • sinx=1cscx\sin x = \frac{1}{\csc x}

  • cosx=1secx\cos x = \frac{1}{\sec x}

  • tanx=1cotx\tan x = \frac{1}{\cot x}

  • cscx=1sinx\csc x = \frac{1}{\sin x}

  • secx=1cosx\sec x = \frac{1}{\cos x}

  • cotx=1tanx\cot x = \frac{1}{\tan x}

Double Angle Identities

  • sin(2x)=2sinxcosx\sin{(2x)} = 2 \sin x \cos x

  • cos(2x)=cos2xsin2x=2cos2x1=12sin2x\cos{(2x)} = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x

  • tan(2x)=2tanx1tan2x\tan{(2x)} = \frac{2 \tan x}{1 - \tan^2 x}

You can use trigonometric substitution with the above identities to help solve tricky integrals involving trigonometric functions.

For example, let’s evaluate \int_0^{\pi} \sin^2 x \, \dx. We’ll first use the double angle identity \cos{(2x)} = 1 - 2\sin^2 x, rearranging it to find that \sin^2 x = \frac{1}{2}(1-\cos{(2x)}. Then, we’ll use the trigonometric integral rule for cosine with an inner coefficient and integrate as usual.

0πsin2xdx=0π12(1cos(2x)dx\int_0^{\pi} \sin^2 x \, dx = \int_0^{\pi} \frac{1}{2}(1-\cos{(2x)} \, dx
=[12(xsin(2x))]0π= [\frac{1}{2} (x - \sin{(2x)})] \Big|_0^{\pi}
=[12x14sin(2x)]0π= [\frac{1}{2}x - \frac{1}{4}\sin{(2x)}] \Big|_0^{\pi}
=π2= \frac{\pi}{2}

More Calculus Practice

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